<p><strong>29.</strong> If a chord of the circle \(x^2 + y^2 - 4x - 2y - c = 0\) is trisected at the points \((1/3, 1/3)\) and \((8/3, 8/3)\), then the radius of the circle will be:</p>
Step-by-Step Solution
Key Concept: A chord trisected by two points means those points divide the chord into three equal parts. Use the property that the perpendicular from the center to a chord bisects the chord, combined with the midpoint and distance relationships of the trisection points.
<p><strong>Step 1:</strong> Let the two trisection points be P(1/3, 1/3) and Q(8/3, 8/3). If these trisect chord AB, then AP = PQ = QB.</p><p><strong>Step 2:</strong> Find distance PQ: PQ = √[(8/3 - 1/3)² + (8/3 - 1/3)²] = √[(7/3)² + (7/3)²] = √(98/9) = 7√2/3</p><p><strong>Step 3:</strong> Since PQ is 1/3 of total chord, the chord length AB = 3 × (7√2/3) = 7√2</p><p><strong>Step 4:</strong> The direction of chord is along vector PQ = (7/3, 7/3), so slope = 1. The perpendicular from center has slope = -1.</p><p><strong>Step 5:</strong> Midpoint of AB lies on the perpendicular bisector. Since P and Q trisect AB symmetrically, midpoint M of AB is at the midpoint of PQ: M = ((1/3 + 8/3)/2, (1/3 + 8/3)/2) = (3/2, 3/2)</p><p><strong>Step 6:</strong> Rewrite circle as (x-2)² + (y-1)² = 4 + 1 + c = 5 + c. Center is C(2, 1).</p><p><strong>Step 7:</strong> Distance from center C(2,1) to chord midpoint M(3/2, 3/2): d = √[(2-3/2)² + (1-3/2)²] = √(1/4 + 1/4) = √(1/2) = 1/√2</p><p><strong>Step 8:</strong> Using chord property: r² = d² + (half-chord)² = 1/2 + (7√2/6)² = 1/2 + 49/18 = 9/18 + 49/18 = 58/18 = 29/9</p><p><strong>Step 9:</strong> Therefore r = √(29/9) = √29/3, so r² = 29/9, giving radius = <strong>√29/3</strong></p><p>∴ Answer: C</p>
Correct Answer: C